12.
Na2CO3+H2SO4->Na2SO4+H2O+CO2
............. 0,5 ............. ......... 0,5
CO2+2KOH->K2CO3+H2O
x 2x x
CO2+KOH->KHCO3
y y y
mKOH=98.40/100=39,2g
nKOH=39,2/56=0,7mol
Có:
2x+y=0,7
138x+100y=57,6
=>x=0,2mol; y=0,3mol
mK2CO3=138.0,2=27,6g
mKHCO3=57,6-27,6=30g
b.
nCO2=x+y=0,2+0,3=0,5mol
CMddH2SO4=0,5/0,2=2,5M
8. Hoàn thành sơ đồ chuyển hóa sau:
Mg \(\underrightarrow{\left(1\right)}\) MgO \(\underrightarrow{\left(2\right)}\) MgCl2 \(\underrightarrow{\left(3\right)}\) Mg(OH)2 \(\underrightarrow{\left(4\right)}\) MgO \(\underrightarrow{\left(5\right)}\) MgSO4 \(\underrightarrow{\left(6\right)}\) MgCO3 \(\underrightarrow{\left(7\right)}\) MgO
\(\left(1\right)2Mg+O_2\underrightarrow{t^o}2MgO\)
\(\left(2\right)MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(\left(3\right)MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
\(\left(4\right)Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)
\(\left(5\right)MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(\left(6\right)MgSO_4+Na_2CO_3\rightarrow MgCO_3+Na_2SO_4\)
\(\left(7\right)MgCO_3\underrightarrow{t^o}MgO+CO_2\uparrow\)
9.
Đổi: 100ml = 0,1l
\(n_{HCl\left(1\right)}=n_{HCl\left(2\right)}=C_M.V_{dd}=1.0,1=0,1\left(mol\right)\)
PTHH: 2HCl + CaCO3 \(\rightarrow\) CaCl2 + H2O + CO2\(\uparrow\) (1)
Theo pt: .. 2 .......... 1 ............. 1 ........ 1 ......... 1 ........ (mol)
Theo đề: . 0,1 ..... 0,05 ........ 0,05 ... 0,05 .... 0,05 ..... (mol)
\(n_{CaCO_3}=\dfrac{m}{M}=\dfrac{20}{100}=0,2\left(mol\right)\)
So sánh \(\dfrac{n_{HCl_{đề}}}{n_{HCl_{pt}}}< \dfrac{n_{CaCO_{3_{đề}}}}{n_{CaCO_{3_{pt}}}}\left(\dfrac{0,1}{2}< 0,2\right)\)
=> CaCO3 dư, tính theo nHCl
=> Dung dịch sau phản ứng gồm CaCl2, H2O và CaCO3 dư.
\(m_{CaCO_{3_{dư}}}=n.M=\left(0,2-0,05\right).100=15\left(g\right)\)
\(m_{CaCl_2}=n.M=0,05.111=5,55\left(g\right)\)
\(m_{H_2O}=n.M=0,05.18=0,9\left(g\right)\)
\(m_{CO_2}=n.M=0,05.44=2,2\left(g\right)\)
\(\Rightarrow m_{dd_{spứ\left(1\right)}}=m_{CaCO_{3_{dư}}}+m_{CaCl_2}+m_{H_2O}-m_{CO_2}\)
\(\Rightarrow m_{dd_{spứ\left(1\right)}}=15+5,55+0,9-2,2=19,25\left(g\right)\) (*)
PTHH: 2HCl + BaCO3 \(\rightarrow\) BaCl2 + H2O + CO2\(\uparrow\) (2)
Theo pt: . 2 .......... 1 ............. 1 ......... 1 ......... 1 ........ (mol)
Theo đề: 0,1 ...... 0,05 ......... 0,05 ... 0,05 .... 0,05 ..... (mol)
\(n_{BaCO_3}=\dfrac{m}{M}=\dfrac{20}{197}\left(mol\right)\)
So sánh \(\dfrac{n_{HCl_{đề}}}{n_{HCl_{pt}}}< \dfrac{n_{BaCO_{3_{đề}}}}{n_{BaCO_{3_{pt}}}}\left(\dfrac{0,1}{2}< \dfrac{20}{197}\right)\)
=> BaCO3 dư, tính theo nHCl
=> Dung dịch sau phản ứng gồm BaCl2, H2O và BaCO3 dư.
\(m_{BaCO_{3_{dư}}}=n.M=\left(\dfrac{20}{197}-0,05\right).197=10,15\left(g\right)\)
\(m_{BaCl_2}=n.M=0,05.208=10,4\left(g\right)\)
\(m_{H_2O}=n.M=0,05.18=0,9\left(g\right)\)
\(m_{CO_2}=n.M=0,05.44=2,2\left(g\right)\)
\(\Rightarrow m_{dd_{spứ\left(2\right)}}=m_{BaCO_{3_{dư}}}+m_{BaCl_2}+m_{H_2O}-m_{CO_2}\)
\(\Rightarrow m_{dd_{spứ\left(2\right)}}=10,15+10,4+0,9-2,2=19,25\left(g\right)\) (**)
Từ (*) và (**) suy ra sau khi kết thúc phản ứng, đĩa cân ở vị trí cân bằng.